You have found the following ages (in years) of all 5 porcupines at your local zoo: $ 16,\enspace 10,\enspace 5,\enspace 7,\enspace 13$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{16 + 10 + 5 + 7 + 13}{{5}} = {10.2\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $16$ years $5.8$ years $33.64$ years $^2$ $10$ years $-0.2$ years $0.04$ years $^2$ $5$ years $-5.2$ years $27.04$ years $^2$ $7$ years $-3.2$ years $10.24$ years $^2$ $13$ years $2.8$ years $7.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{33.64} + {0.04} + {27.04} + {10.24} + {7.84}} {{5}} $ $ {\sigma^2} = \dfrac{{78.8}}{{5}} = {15.76\text{ years}^2} $ The average porcupine at the zoo is 10.2 years old. The population variance is 15.76 years $^2$.